# Find a point on t

Given:

The curve y = x3 – 2x2 – 2x and a line y = 2x – 3

First, we will find The Slope of tangent

y = x3 – 2x2 – 2x (x3) – (2x2) – (2x) = 3x3 – 1 (x2 – 1) – 2 x1 – 1 = 3x2 – 4x – 2 ...(1)

y = 2x – 3 is the form of equation of a straight line y = mx + c,where m is the The Slope of the line.

so the The Slope of the line is y = 2 (x) – 3

Thus, The Slope = 2. ...(2)

From (1) & (2)

3x2 – 4x – 2 = 2

3x2 – 4x = 4

3x2 – 4x – 4 = 0

We will use factorization method to solve the above Quadratic equation.

3x2 – 6x + 2x – 4 = 0

3x(x – 2) + 2(x – 2) = 0

(x – 2)(3x + 2) = 0

(x – 2) = 0 & (3x + 2) = 0

x = 2 or

x = Substitute x = 2 & x = in y = x3 – 2x2 – 2x

when x = 2

y = (2)3 – 2 (2)2 – 2 (2)

y = 8 – (2 4) – 4

y = 8 – 8 – 4

y = – 4

when x = y = ( )3 – 2 ( )2 – 2 ( )

y = ( ) – 2 ( ) + ( )

y = ( ) – ( ) + ( )

taking lcm

y = y = y = Thus, the points are (2, – 4) & ( , )

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